Forum:Feeding FGH into FGH
I'm a newbie in googology, and I have some questions about FGH and ordinals. FGH looks like this: f_\alpha(n) while \alpha can be an integer or an ordinal. what I want to do is to feed it into it self: f_{f_\alpha(\omega)}(n) but here's some problem about this: 1.how to define f_\alpha(\omega) when \alpha is an limit ordinal? fundamental sequence is totally useless. 2.what's the limit of this? is that the fix point of f_\alpha(\omega)=\alpha ? if so, the ordinal \alpha is probably \Gamma_0 I guess,which is not even as powerful as the original FGH!!! 3. \Gamma_0 can be expressed as {\{\omega,\omega,1,2}\} in BEAF. it seems like that it is possible to range \Gamma_0 by using f_\alpha(\omega) . what's the problem between 2 and 3? —Preceding unsigned comment added by D57799 (talk • ) :Hey, welcome to the wiki. Feeding FGH back into itself is actually considered before, though it is a good idea anyways. :Defining this is indeed quite difficult, but you can see my definition here. It turns out, that, under this definition, it actually reaches \(\vartheta(\Omega_\omega)\). Defining fixed points of this using \(\Omega\), just as happens in \(\vartheta\) or \(\psi\), is actually very, very powerful. :The problem for defining ordinals in BEAF is that you need a definition, otherwise things will go wrong. Wythagoras (talk) 06:51, September 28, 2014 (UTC) Thanks. It's really helpful. I think that probably exceeded any ordinal collapsing function used by now.D57799 (talk) 07:23, September 28, 2014 (UTC) ::Wythagoras, I'm afraid your analysis is incorrect. ::As you say, \(f_\omega(\omega 2) = \sup(f_0 (\omega 2), f_1(\omega_2),f_2(\omega 2) \ldots)\). Since \(f_n(\omega 2) < \varphi(\omega,0)\) for all finite \(n\), it is impossible for \(f_\omega(\omega 2)\) to be greater than \(\varphi(\omega, 0)\), and in fact it is equal to \(\varphi(\omega, 0)\). The same argument holds for all \(f_\omega(\alpha)\) with \(\alpha < \varphi(\omega, 0)\), so we have that for all \(2 \omega \le \alpha < \varphi(\omega,0)\), \(f_\omega(\alpha) = \varphi(\omega,0)\). More generally, for all infinite \(\alpha\), \(f_\omega(\alpha)\) will be the smallest ordinal of the form \(\varphi(\omega,\beta)\) that is larger than \(\alpha\). ::For \(f_{\omega+1}(\alpha) = f_{\omega}^{\alpha}(\alpha)\), we start at \(\alpha\) and apply \(f_\omega \alpha\) times, and each time it goes up to the next ordinal of the form \(\varphi(\omega, \beta)\). So we will usually get \(f_{\omega+1}(\alpha) = \varphi(\omega, \alpha)\). (There is a slight difference near epsilon numbers.) ::If you continue this analysis, you should get that \(f_{\alpha+1}(\beta)\) is about \(\varphi(\alpha, \beta)\), and so for \(\alpha, \beta < \Gamma_0, f_\alpha(\beta) < \Gamma_0\). ::My version of the hierarchy is almost the same, but the definition is somewhat simpler: ::\(f_0(\beta) = \beta + 1\) ::\(f_{\alpha+1}(\beta) = f_\alpha^{\beta}(\beta)\) ::for limit \(\alpha, f_\alpha(\beta) = \sup(\gamma < \alpha) \{f_\gamma(\beta)\}\) ::\(f^{\lambda+1}(\beta) = f(f^{\lambda}(\beta))\) ::for limit \(\lambda, f^\lambda(\beta) = \sup(\gamma < \lambda) \{f^\gamma(\beta)\}\) ::No need to invoke fundamental sequences. Deedlit11 (talk) 07:35, September 28, 2014 (UTC) ::I shall try to redefine it. Wythagoras (talk) 08:22, September 28, 2014 (UTC) :D57799: Actually, ordinal collapsing functions go very far beyond \(\vartheta(\Omega_\omega)\). You can see for example my blog posts "Ordinal Notations I-VI". In fact, OCF's go much further than what I describe, but I don't understand the larger ones very well. Deedlit11 (talk) 17:13, September 28, 2014 (UTC) I got it wrong about the fix point of f_\alpha(\omega)=\alpha . I think I should be more familiar about how ordinals work.--D57799 (talk) 10:54, September 29, 2014 (UTC) What is the exact point that SGH(n) equals FGH(n)? it seems that g_\alpha(\omega)=\alpha when \alpha is an ordinal. So the fix point of \alpha=f_\alpha(\omega)=f_{g_\alpha(\omega)}(\omega) is actually where the f_\alpha(\omega)=g_\alpha(\omega) , the point that SGH(n) equals FGH(n). the page of SGH shows that will be \psi_0(\Omega_\omega) is that correct?D57799 (talk) 13:07, September 29, 2014 (UTC) The problem with Deedlit's definition of FGH is that it's designed to work with ordinals - as you can check, for example, using Deedlit's definition \(f_\omega(2)=\omega\), which would never happen with standard FGH, where result is always finite. As you point out, in Deedlit's version the meeting point is very soon, but in standard finite version it appears much later, at \(\psi(\Omega_\omega)\). I also believe that it isn't real equality, but it just means that growth rates are comparable. LittlePeng9 (talk) 13:21, September 29, 2014 (UTC) What is the formal way to do f_3(\omega) ? obviously it's f_2^\omega(\omega) . but I always get \omega^\omega instead of \epsilon_0 , I get it at f_4(\omega) Besides, f_2(\omega)=2^\omega\omega=\omega^2 right? :\(f_0(\alpha) = \alpha+1\) :\(f_1(\alpha) = \alpha*2\) :\(f_2(\alpha) = \alpha * 2^\alpha\) :\(f_2(\omega) = \omega * 2^\omega = \omega^2\) :\(f_2(\omega^2) = \omega^2 * 2^{\omega^2} = \omega^2 * (2^\omega)^\omega) = \omega^2 * \omega^\omega = \omega^\omega\) :\(f_2(\omega^\omega) = \omega^\omega * 2^{\omega^\omega} = \omega^\omega * 2^{\omega * \omega^\omega} = \omega^\omega * (2^\omega)^{\omega^\omega} = \omega^{\omega^\omega}\) :\(f_2(\omega^{\omega^\omega}) = \omega^{\omega^\omega} * 2^{\omega^{\omega^\omega}} = \omega^{\omega^\omega}*(2^\omega)^{\omega^{\omega^\omega}} = \omega^{\omega^{\omega^\omega}}\) :and so on. You can see that the limit is \(\varepsilon_0\). :More generally, \(f_2^\omega(\alpha) = \lim \{\alpha^{\alpha^{\alpha^\cdots}}\}\), which is the smallest epsilon number larger than \(\alpha\). :So, \(f_3(\omega \alpha)\) is the \(\alpha\)th epsilon number larger than \(\omega \alpha\), which will usually be \(\varepsilon_\alpha\) except near fixed points of the \(\varepsilon\) function. Deedlit11 (talk) 15:04, September 30, 2014 (UTC) :Oh,I see. f_2(\omega^2)=(\omega^2)2^{(\omega^2)} , not just (\omega^2)^2 .D57799 (talk) 11:00, October 1, 2014 (UTC) I'm trying to fit that into FGH. I found out that in the all-integer case, it is amazingly weak.: f_{f_{f_{...}(n)}(n)}(n) > 2\uparrow^{2\uparrow^{2\uparrow^{...}n}n}n Which is merely between f_{\omega+1}(n) and f_\omega(n) , and it is much smaller than f_{\Gamma_0}(n) , which is actually around f_{f_{\omega+1}(\omega)}(n) . It is surely smaller than chained arrow's f_{\omega^2}(n) , and linear BEAF's f_{\omega^\omega}(n) , but I'm quite certain about that f_{f_{f_{...}(\omega)}(\omega)}(n) will grow much faster than f_{\omega^\omega}(n) . Besides, Why is the comparison between FGH and BEAF deleted a great part? It said that BEAF is "nonfunctional" after \varepsilon_0 . Does that mean it's not possible to compare or something else? D57799 (talk) 05:23, October 4, 2014 (UTC) : For quite a long time it was known that what Bowers has left us from the definition of BEAF is not enough to make sense out of his notation beyond e0. There have been ideas on how to extend this, I believe we had a solid idea up to at least \(\varphi(\omega,0)\), but since there turned out to be multiple ways to even go there, we decided to mark it as "nonfunctional" beyond that point. We might put it back there if we find a single way to define BEAF up to some reasonable point. LittlePeng9 (talk) 05:51, October 4, 2014 (UTC)